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HDOJ 1018 Big Number（大数位数公式）

Problem Description In many applications very large integers numbers are required. Some of these applications are using keys for secure transmission of data, encryption, etc. In this problem you are given a number, you have to determine the number of…

计蒜客 31452 - Supreme Number - [简单数学][2018ICPC沈阳网络预赛K题]

题目链接:https://nanti.jisuanke.com/t/31452 A prime number (or a prime) is a natural number greater than $1$ that cannot be formed by multiplying two smaller natural numbers. Now lets define a number $N$ as the supreme number if and only if each number m…

hdoj--1018--Big Number(简单数学)

Big Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 32410    Accepted Submission(s): 15175 Problem Description In many applications very large integers numbers are required. Some of thes…

SDUT 3258 Square Number 简单数学

和上一题一样,把平方因子除去,然后对应的数就变成固定的 #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> using namespace std; typedef long long LL; ; const int INF=0x3f3f3f3f; ],cnt; void getprime(){ ]; memset(v,,sizeof(v)); ;i*i&…

SDUT 3257 Cube Number 简单数学

把所有数的立方因子除去,那么一个数可以和它组成立方的数是确定的,统计就行 #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <cmath> using namespace std; typedef long long LL; ; const int INF=0x3f3f3f3f; ],cnt; void getprime()…

HDU 5073 Galaxy (2014 Anshan D简单数学)

HDU 5073 Galaxy (2014 Anshan D简单数学) 题目链接http://acm.hdu.edu.cn/showproblem.php?pid=5073 Description Good news for us: to release the financial pressure, the government started selling galaxies and we can buy them from now on! The first one who bought…

HDOJ 1501 Zipper 【简单DP】

HDOJ 1501 Zipper [简单DP] Problem Description Given three strings, you are to determine whether the third string can be formed by combining the characters in the first two strings. The first two strings can be mixed arbitrarily, but each must stay in i…

水题 HDOJ 4727 The Number Off of FFF

题目传送门 /* 水题:判断前后的差值是否为1,b[i]记录差值,若没有找到,则是第一个出错 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <string> #include <cmath> using namespace std; ; const int INF = 0x3f3f3f3f; int a…

洛谷试炼场-简单数学问题-P1403 [AHOI2005]-因数

洛谷试炼场-简单数学问题 P1403 [AHOI2005]约数研究 Description 科学家们在Samuel星球上的探险得到了丰富的能源储备,这使得空间站中大型计算机"Samuel II"的长时间运算成为了可能.由于在去年一年的辛苦工作取得了不错的成绩,小联被允许用"Samuel II"进行数学研究. 小联最近在研究和约数有关的问题,他统计每个正数N的约数的个数,并以f(N)来表示.例如12的约数有1.2.3.4.6.12.因此f(12)=6.下表给出了一些f…

洛谷试炼场-简单数学问题-P1045 麦森数-高精度快速幂

洛谷试炼场-简单数学问题 B--P1045 麦森数 Description 形如2^P−1的素数称为麦森数,这时P一定也是个素数.但反过来不一定,即如果PP是个素数,2^P-1 不一定也是素数.到1998年底,人们已找到了37个麦森数.最大的一个是P=3021377P=3021377,它有909526位.麦森数有许多重要应用,它与完全数密切相关. 任务:从文件中输入PP(1000<P<31000001000<P<3100000),计算2^P-1 的位数和最后500位数字(用十进制高…

洛谷试炼场-简单数学问题-P1088 火星人

洛谷试炼场-简单数学问题 A--P1088 火星人 Description 人类终于登上了火星的土地并且见到了神秘的火星人.人类和火星人都无法理解对方的语言,但是我们的科学家发明了一种用数字交流的方法.这种交流方法是这样的,首先,火星人把一个非常大的数字告诉人类科学家,科学家破解这个数字的含义后,再把一个很小的数字加到这个大数上面,把结果告诉火星人,作为人类的回答. 火星人用一种非常简单的方式来表示数字――掰手指.火星人只有一只手,但这只手上有成千上万的手指,这些手指排成一列,分别编号为1,2,…

简单数学算法demo和窗口跳转,关闭,弹框

 简单数学算法demo和窗口跳转,关闭,弹框demo <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="zh-cn&quo…

hdu 2200 Eddy's AC难题（简单数学。。）

题意: N个人,每个人AC的题数都不一样. Eddy想从中选出一部分人(或者全部)分成两组.必须满足第一组中的最小AC数大于第二组中的最大AC数. 问共有多少种不同的选择方案. 思路: 简单数学.. 代码: ll C(int n,int x){ ll ans=1; rep(i,1,x){ ans = ans*(n+1-i)/i; } return ans; } int main(){ int n; while(cin>>n){ ll ans = 0; rep(i,2,n){ ans += (C…

hdu 1018 Big Number 数学结论

Big Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34084    Accepted Submission(s): 16111 Problem Description In many applications very large integers numbers are required. Some of these…

HDU 1018 Big Number 数学题解

Problem Description In many applications very large integers numbers are required. Some of these applications are using keys for secure transmission of data, encryption, etc. In this problem you are given a number, you have to determine the number of…

【HDOJ】1018 Big Number

数学题,还是使用log避免大数,但是不要忘记需要+1,因为0也是1位,log(100)= 2,但却是3位. #include <stdio.h> #include <math.h> int main() { int case_n, n; int i; double sum; scanf("%d", &case_n); while (case_n--) { scanf("%d", &n); sum = 1.0f; ; i<…

hdu 1018:Big Number（水题）

Big Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 21106    Accepted Submission(s): 9498 Problem Description In many applications very large integers numbers are required. Some of these…

POJ 1844 Sum【简单数学】

链接: http://poj.org/problem?id=1844 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=29256#problem/D Sum Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 9795   Accepted: 6406 Description Consider the natural numbers from 1 to N. By a…

HDOJ 3709 Balanced Number

数位DP... Balanced Number Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 1337    Accepted Submission(s): 583 Problem Description A balanced number is a non-negative integer that can be balanced…

HDOJ 4937 Lucky Number

当进制转换后所剩下的为数较少时(2位.3位),相应的base都比較大.能够用数学的方法计算出来. 预处理掉转换后位数为3位后,base就小于n的3次方了,能够暴力计算. . .. Lucky Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 521    Accepted Submission(s): 150 Probl…

HDU 1018-Big Number（数学）

Big Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 26383    Accepted Submission(s): 12006 Problem Description In many applications very large integers numbers are required. Some of thes…

HDU 5166 Missing number 简单数论

Missing number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) [Problem Description] There is a permutation without two numbers in it, and now you know what numbers the permutation has. Please find the two numbers…

PAT 1019 General Palindromic Number[简单]

1019 General Palindromic Number (20)(20 分) A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers. Altho…

PAT 1144 The Missing Number[简单]

1144 The Missing Number(20 分) Given N integers, you are supposed to find the smallest positive integer that is NOT in the given list. Input Specification: Each input file contains one test case. For each case, the first line gives a positive integer…

B. Div Times Mod Round #528 (Div. 2)【简单数学】

一.题面 题目链接 二.分析 一个简单的数学题目,这里首先要把x分解了看 $x = kd + c$ 这样原问题中的n就变成了 $n = dc$ 上面这个式子中,c因为是x除k取余得到的,那么可以肯定 $c < k$ 有了这个式子,就可以直接暴力去试满足条件的c,并且最小的d就可以满足x的最小值. 三.AC代码 #include <bits/stdc++.h> using namespace std; #define INF 1e6+4 int main() { int K, N; //f…

HDOJ(1018)

Big Number Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34743    Accepted Submission(s): 16478 Problem Description In many applications very large integers numbers are required. Some of these…

PAT甲级——1104 Sum of Number Segments (数学规律、自动转型)

本文同步发布在CSDN:https://blog.csdn.net/weixin_44385565/article/details/90486252 1104 Sum of Number Segments (20 分)   Given a sequence of positive numbers, a segment is defined to be a consecutive subsequence. For example, given the sequence { 0.1, 0.2, 0.…

poj1142.Smith Number（数学推导）

Smith Number Time Limit: 1 Sec  Memory Limit: 64 MB Submit: 825  Solved: 366 Description While skimming his phone directory in 1982, Albert Wilansky, a mathematician of Lehigh University,noticed that the telephone number of his brother-in-law H. Smit…

HDU 1018 Big Number （数学题）

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1018 解题报告:输入一个n,求n!有多少位. 首先任意一个数 x 的位数 = (int)log10(x) + 1; 所以n!的位数 = (int)log10(1*2*3*.......n) + 1; = (int)(log10(1) + log10(2) + log10(3) + ........ log10(n)) + 1; #include<cstdio> #include<cstrin…

Codeforces 665D Simple Subset [简单数学]

题意: 给你n个数,让你从中选一个子集要求子集中的任何两个数相加都是质数. 思路: 一开始把自己坑了,各种想,后来发现一个简单的性质,那就是两个数相加的必要条件是这两个数之中必定一个奇数一个偶数,(除了含有1 集合以外,1+1等于2也是质数). 考虑两种情况,有1存在和1不存在这两种. 很显然1存在的情况下,所有的1都可以同时在集合中出现,要想集合最大不能加奇数,只能加偶数,那么我们看原始集合中是否有偶数加一是素数. 不考虑1的情况下,这样的子集最大是2,只有存在一个奇数一个偶数相加是质数的情况…